Chapter 18: Problem 66
The equation of the directrix of the parabola \(y^{2}+4 y+\) \(4 x+2=0\) is: (A) \(x=-1\) (B) \(x=1\) (C) \(x=-\frac{3}{2}\) (D) \(x=\frac{3}{2}\)
Short Answer
Expert verified
The equation of the directrix is \( x = \frac{3}{2} \) (Option D).
Step by step solution
01
Rewrite Equation in Standard Form
First, we need to rearrange the given equation to match the standard form of a parabola equation. The given equation is \( y^2 + 4y + 4x + 2 = 0 \). Let's rearrange it: \( y^2 + 4y = -4x - 2 \).
02
Complete the Square for the y-terms
To simplify the equation, we complete the square for the \( y \) terms. Start with \( y^2 + 4y \). Add and subtract \(4\) (i.e., \( (\frac{4}{2})^2 = 4\)) inside the equation: \( y^2 + 4y + 4 - 4 = -4x - 2 \). This becomes \( (y + 2)^2 - 4 = -4x - 2 \).
03
Simplify the Equation
Now convert the equation into standard form by moving constants: \( (y + 2)^2 = -4x + 2 \). Simplify the constants to \( (y + 2)^2 = -4(x - \frac{1}{2}) \).
04
Identify Parabola Parameters
This equation \( (y+2)^2 = -4(x-\frac{1}{2}) \) represents a parabola that opens to the left. The vertex form \( (y-k)^2 = 4a(x-h) \) shows that the vertex \((h, k)\) is \((\frac{1}{2}, -2)\) and \( 4a = -4 \) implies \( a = -1 \).
05
Find the Equation of the Directrix
The equation for the directrix of a parabola in the form \( (y-k)^2 = 4a(x-h) \) is \( x = h - a \). Since \( h = \frac{1}{2} \) and \( a = -1 \), the directrix is \( x = \frac{1}{2} - (-1) = \frac{1}{2} + 1 = \frac{3}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of a Parabola
Parabolas are symmetrical curves and one of the most recognizable shapes in mathematics. They have a special equation that describes their shape and position on a coordinate plane. The standard form of a parabola centered around the x-axis is:
For example, in the problem given, we have a horizontal parabola because the x is on one side by itself after manipulation. Rewriting the initial equation: \[ y^2 + 4y + 4x + 2 = 0 \] aids in determining this as a horizontal parabola when compared to the general formula.
- Vertical parabolas: \( y = ax^2 + bx + c \)
- Horizontal parabolas: \( x = ay^2 + by + c \)
For example, in the problem given, we have a horizontal parabola because the x is on one side by itself after manipulation. Rewriting the initial equation: \[ y^2 + 4y + 4x + 2 = 0 \] aids in determining this as a horizontal parabola when compared to the general formula.
Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square form. This technique is particularly useful for solving quadratic equations and rewriting equations of parabolas.
Here's a simple way to understand completing the square:
Completing the square makes it simpler to transform the equation into vertex form and identify critical points of the parabola.
Here's a simple way to understand completing the square:
- First, identify the quadratic terms, such as \(y^2 + 4y\).
- Next, take half of the coefficient of the linear term \(b\), square it, and add and subtract this square within the equation to preserve equality.
- In our example: \((\frac{4}{2})^2 = 4\).
Completing the square makes it simpler to transform the equation into vertex form and identify critical points of the parabola.
Parabola Vertex Form
The vertex form is a way of expressing the equation of a parabola to easily identify its vertex and direction. It looks like this for a parabola opening horizontally:\[ (y-k)^2 = 4a(x-h) \]Where:
- \((h, k)\) is the vertex of the parabola.
- \(a\) is a coefficient that determines the direction and width of the parabola.
- The vertex of the parabola is \((\frac{1}{2}, -2)\).
- The parabola opens to the left since \(a = -1\) is negative.